that is, [itex]b[/itex] is the negative of the first column. Written as it is above, [itex]A[/itex] has zero determinant and the determinant formed when the[itex] k^{th}[/itex] column of [itex]A[/itex] is substituted by the vector [itex]b[/itex] is clearly zero as well. A theorem says that in this case the system has infinite solutions. If one reduces the system to reduced row-echelon form the solutions can be parameterized as, for example, [itex]x_3=t,x_2=5t/9,x_1=t−1[/itex]. An immediate solution by inspection is [itex]x=(−1,0,0)^T[/itex] which one obtains letting [itex]t=0[/itex].

But let's give another value of [itex]t[/itex], for example, [itex]t=1[/itex] which gives [itex]x=(0,5/9,1)^T[/itex]. This is one of the parameterized solutions and yet it does not satisfy the original system. It does, however, satisfy the reduced system obtained from the original by gaussian elimination and should be equivalent, i.e.,

[tex]
\begin{cases}
x_1-x_3 &=-1\\
9x_2-5x_3 & = 0
\end{cases}
[/tex]

But shouldn't my parameterized solution satisfy both original and reduced systems, no matter what? Yet, the only satisfying solution for the original system seems to be [itex]x=(−1,0,0)^T[/itex]. What am I not seeing here?

is contained in the kernel of the mentioned matrix, and all other elements of the kernel can be obtained by scaling this. It seems you have attempted to scale this by a factor [itex]\frac{t}{9}[/itex] with a real coefficient [itex]t[/itex], but you have made a mistake with the [itex]x_1[/itex] component.